c 1997 International Press
°
005
ASIAN J. MATH.
Vol. 1, No. 2, pp. 272–292, June 1997
THE REGULARITY OF Lp -SCALING FUNCTIONS∗
KA-SING LAU† and MANG-FAI MA‡
Abstract. The existence of Lp -scaling function and the Lp -Lipschitz exponent have been studied in [Ji] and [LW] and a criterion is given in terms of a series of product of matrices. In this paper
we make some further study of the criterion. In particular we show that for p an even integer or for
all p ≥ 1 in some special cases, the criterion can be simplified to a computationally efficient form.
1. Introduction. The solution f of a 2-scale dilation equation
(1.1)
f (x) =
N
X
cn f (2x − n),
x∈R
n=0
is called a scaling function. This class of functions has been studied in detail in recent literature in connection with wavelet theory [D] and constructive approximations
[DGL]. The question of existence of continuous, L1 and L2 solutions was treated in
Daubechies [D], Daubechies and Lagarias [DL1], Collela and Heil [CH], Eirola [E], Heil
[H], and Micchelli and Prautzsch [MP]. The regularity of such solutions was studied,
in addition to the above papers, in Cohen and Daubechies [CD], Daubechies and Lagarias [DL2,3], Herve [He], Lau, Ma and Wang [LWM] and Villemos [V1,2]. Also the
existence of Lp -solutions has been characterized by Lau and Wang in [LW] and Jia[Ji].
In this paper, we will adopt the previous notations as in [CH], [DL1] and [LW].
Let T0 = [c2i−j−1 ]1≤i,j≤N and T1 = [c2i−j ]1≤i,j≤N be the associated matrices of the
coefficients {cn }, i.e.,




c0 0 0 . . .
0
c1 c0 0 . . . 0
0 
 c3 c2 c1 . . . 0 
 c2 c1 c0 . . .




c5 c4 c3 . . . 0  .
c
c
c
.
.
.
0


4
3
2
, T1 = 
T0 = 



.
.
.
.
.
.
.
.
 ..
 ..
..
.. . . .
.. 
..
.. . . .
.. 
0 0 0 . . . cN
0 0 0 . . . cN −1
PN
It is known that
n=0 cn = 2, then 2 is always an eigenvalue of (T0 + T1 ). FurP if P
thermore, if
c2n =
c2n+1 = 1, then 1 is an eigenvalue of both T0 and T1 . Let v
be a 2-eignevector of (T0 + T1 ) (which means a right eigenvector associated with the
eigenvalue 2) and let
ṽ := (T0 − I)v = (I − T1 )v.
In [LW] the following theorem was proved:
PN
Theorem A. Suppose 1 ≤ p < ∞ and n=0 cn = 2. Then equation (1.1) has a
nonzero compactly supported Lp -solution (notation: Lpc -solution) if and only if there
exists a 2-eigenvector v of (T0 + T1 ) satisfying
1 X
lim n
kTJ ṽkp = 0.
n→∞ 2
|J|=n
∗
Received March 20, 1997; accepted for publication June 25, 1997.
Department of Mathematics, The Chinese University of Hong Kong, Shatin, Hong Kong
(kslau@math.cuhk.edu.hk).
‡ Department of Mathematics, University of Pittsburgh, Pittsburgh, PA 15260.
†
272
the regularity of Lp -scaling functions
273
In [Ji], Jai studied the same existence question by means of a ”hat’ function and
obtained a similar criterion independently.
P
P Furthermore he showed that the existence
of an Lp -solution implies that
c2n = c2n+1 = 1.
In this paper we will consider the regularity of the solution. We use the Lp Lipschitz exponent to describe the regularity. It is defined by
Lipp (f ) = lim inf
+
h→0
ln k∆h f kp
,
ln h
where ∆h f (x) = f (x + h) − f (x). It is well known that for 1 ≤ p < ∞, if
1
lim sup k∆h f kp < ∞
h→0+ h
(which implies Lipp (f ) = 1), then f 0 exists a.e. and is in Lp and f is the indefinite
integral of f 0 . Recall that the q-Sobolev exponent of a function f is defined as
Z
sup{α : (1 + |ξ|qα )|fˆ(ξ)|q dξ < ∞}.
For p = q = 2, the 2-Sobolev exponent equals to the L2 -Lipschitz exponent, and they
are different when p, q 6= 2. In general the Lp -Lipschitz exponent describes the regularity of f more accurately than the Sobolev exponent. The q-Sobolev exponent has
been studied in [He]. The Lp -Lipschitz exponent (in a slightly different terminology)
has been used to investigate the multifractal structure of scaling functions in [DL3]
and [J1,2].
Let H(ṽ) be the complex subspace spanned by { TJ ṽ : J is a multi-index }. (We
use complex scalar because it will be more convenient to deal with the complex eigenvalues and eigenvectors.).
P
P
Theorem B. Suppose that
c2n =
c2n+1 = 1 and either (i) 1 is a simple
PN
eigenvalue of T0 and T1 or (ii) H(ṽ) = { u ∈ CN : i=1 ui = 0 }. Then for f an
Lpc -solution of (1.1), 1 ≤ p < ∞,
P
ln(2−n |J|=n kTJ ṽkp )
.
(1.2)
Lipp (f ) = lim inf
n→∞
p ln(2−n )
We remark that Jia [Ji, Theorem 6.2] proved that the above formula by replacing
kTJ ṽk with kTJ /Hk where H denoted the hyperplane in (ii). Our special perference
on kTJ ṽk is that it allows us to calculate the sum in many cases (see Section 4 and
5). Even though there are some overlaps with Jia’s result, we like to give a full proof
of Theorem B because of completeness and the consistence of the development in the
in the sequel.
To reduce the formula in Theorem B, we only consider the 4-coefficient dilation
equation for simplicity. We show that if in addition c0 + c3 = 1, then
Lipp (f ) =
ln((|c0 |p + |1 − c0 |p )/2)
,
−p ln 2
(Proposition 4.3) and if c0 + c3 = 1/2, then
© ln((|c0 |p + | 12 − c0 |p )/2) ª
Lipp (f ) = min 1,
.
−p ln 2
274
ka-sing lau and mang-fai ma
(Proposition 4.5). Note that the second case contains Daubechies scaling function D4 .
The formula was actually obtained in [DL3] using a different method and assuming
further 1/2 < c0 < 3/4 . For the general case we show that if p is an even integer,
then
X
kTJ ṽkp = aWpn b
(1.3)
|J|=n
for an auxillary (p + 1) × (p + 1) matrix Wp depends only on the coefficients of the
dilation equation and for some vectors a and b ( Proposition 5.1). In particular for
p = 2, the matrix W2 is equivalent to the transition matrix used in [CD], [LW] and
[V] for the existence of L2 -scaling function. By using (1.3) it is easy to show that
the necessary and sufficient condition in Theorem A reduces to ρ(Wp ) < 2 and (1.2)
becomes
ln(ρ(Wp )/2)
Lipp (f ) =
−p ln 2
where ρ(Wp ) is the spectral radius of Wp (Theorem 5.3).
The paper is organized as follows. In Section 2 we include some preliminary results
concerning the eigen-properties of the matrices T0 , T1 and T0 + T1 that we need. We
give a complete proof of Theorem B in Section 3. In Section 4, we will apply Theorem
B to obtain explicit expressions for the two special cases described above. Finally in
Section 5 we construct the matrix Wp in (1.3) and use the spectral radius of Wp to
determine Lipp (f ) when p is an even integer. We also make some remarks concerning
extensions of the construction and discuss some unsolved questions.
2. Preliminaries. Throughout this paper,
otherwise specified, we assume
P unlessP
that 1 ≤ p < ∞, cn ∈ R, c0 , cN 6= 0 and
c2n =
c2n+1 = 1. For any k ≥ 1,
let J = (j1 , . . . , jk ), ji = 0 or 1, be the multi-index and |J| the length of J. Let
IJ = I(j1 ,...,jk ) be the dyadic interval [0.j1 · · · jk , 0.j1 · · · jk + 2−k ). The matrix TJ
represents the product Tj1 · · · Tjk . If v is a 2-eigenvector of (T0 + T1 ), it is clear that
(2.1)
1
1 X
TJ v = k (T0 + T1 )k v = v.
2k
2
|J|=k
For any g ∈ Lp (R) with support in [0, N ], let g : [0, 1] → RN
g(x) = [g(x), g(x + 1), . . . , g(x + (N − 1))]t ,
x ∈ [0, 1)
be the vector-valued function representing g and let
½
T0 g(2x)
if x ∈ [0, 12 ),
Tg(x) =
T1 g(2x − 1) if x ∈ [ 12 , 1).
It is easy to show that f is a solution of (1.1) if and only if f = Tf [DL1]. With no
confusion, we use k · k to denote the Lp -norm of g as well as the vector-valued function
g. Also for a vector u ∈ Rn , kukR will denote the `pN -norm in RN .
Let gI be the average |I|−1 I g(x) dx of g on an interval I.
Proposition 2.1. Let f be an Lpc -solution of (1.1) and v = [f[0,1) , . . . , f[N −1,N ) ]t
be the vector defined by the average of f on the N subintervals. Then
(i) v is a 2-eigenvector of (T0 + T1 ).
the regularity of Lp -scaling functions
275
(ii) Let f0 (x) = v, x ∈ [0, 1), and let fn+1 = Tfn , n = 0, 1, . . . , then
X
fn (x) =
(TJ v)χIJ (x),
x ∈ [0, 1),
|J|=n
P
and kfn+1 − fn kp = 2−n |J|=n kTJ ṽkp .
P
(iii) kf − fn kp ≤ 2cn |J|=n kTJ ṽkp for some c > 0 and fn → f in Lp ([0, 1], RN ).
Proof. The proof of these statements can be found in [LW, Proposition 2.3,
Lemma 2.4, 2.5, and Theorem 2.6]. In particular to prove the last identity in (ii),
we observe that
1 X
kfn+1 − fn kp = n+1
(kTJ (T0 − I)vkp + kTJ (T1 − I)vkp )
2
|J|=n
1 X
= n
kTJ ṽkp .
2
|J|=n
Let
α = lim inf
ln(2−n
P
|J|=n kTJ ṽk
ln(2−n )
n→∞
p
)
.
P
Then α is the rate of convergence of 2−n |J|=n kTJ ṽkp to 0 in the sense that the sum
is of order o(2−βn ) for any β < α. Let H(ṽ) be the subspace (with complex scalar)
spanned by { TJ ṽ : J is a multi-index }.
Lemma 2.2. Under the same conditions and notations as in Proposition 2.1, for
any u ∈ H(ṽ),
P
ln(2−n |J|=n kTJ ukp )
lim inf
≥ α.
n→∞
ln(2−n )
Furthermore equality holds if H(u) = H(ṽ).
Proof. Since H(ṽ) is finite dimensional, it suffices to consider u = TJ 0 ṽ for some
J 0 . Let |J 0 | = k, then
1 X
1 X
1
kTJ ukp = n
kTJ TJ 0 ṽkp ≤ 2k n+k
n
2
2
2
|J|=n
It follows that
ln(2−n
P
|J|=n kTJ uk
ln(2−n )
|J|=n
p
)
X
kTJ ṽkp .
|J|=n+k
P
ln(2−(n+k) |J|=n+k kTJ ṽkp )
ln(2k )
≥
+
,
ln(2−n )
ln(2−(n+k) )
which implies the stated inequality. For the last statement we need only change the
roles of u and ṽ and make use of the inequality we just proved.
Let M = [c2i−j ]1≤i,j≤N −1 , i.e.,


c1 c0 0 . . .
0
0 
 c3 c2 c1 . . .


c5 c4 c3 . . .
0 
M =
 .
..
.. . .
.. 
 ..
.
.
.
. 
0
0
0
. . . cN −1
276
ka-sing lau and mang-fai ma
P
P
be the common submatrix of T0 and T1 . If
c2n =
c2n+1 = 1, then 1 is an
eigenvalue of M and [1, 1, . . . , 1] is the corresponding left 1-eigenvector. Let H =
PN
{ u ∈ CN : i=1 ui = 0 }.
P
P
Lemma 2.3. There exist v0 , v1 ∈
/ H (i.e.,
(v0 )i =
(v1 )i 6= 0) such that
(T0 − I)m v0 = (T1 − I)m v1 = 0 for some m > 0, and
(2.2)
T0 v1 = T1 v0 .
Remark. When m = 1, v0 and v1 are 1-eigenvectors of T0 and T1 respectively.
Proof. Let Eλ = {u ∈ CN −1 : (M − λI)m u = 0 for some m > 0}. Observe that
for λ 6= 1 and for u ∈ Eλ ,
0 = [1, 1, . . . , 1](M − λI)m u = (1 − λ)m
N
−1
X
ui
i=1
P
P
for some m > 0, so that
ui = 0. In view of CN −1 = E1 ⊕ λ6=1 Eλ , there exists
P
a ∈ E1 such that
ai 6= 0. If 1 is a simple eigenvalue of M , dim E1 = 1 and hence
the above a is a 1-eigenvector of M . Let
v0 := [0, a1 , . . . , aN −1 ]t ,
(2.3)
v1 := [a1 , . . . , aN −1 , 0]t .
Then v0 and v1 are 1-eigenvectors of T0 and T1 respectively, and v0 , v1 ∈
/ H. Moreover, by the definitions of T0 and T1 , we have
X
X
(2.4)
(T0 v1 )i =
c2i−j−1 aj =
c2i−j aj+1 = (T1 v0 )i .
so that T0 v1 = T1 v0 . If 1 is not a simple eigenvalue of M , we let m be the smallest
positive integer so that (M − I)m a = 0. Define a(1) = a, · · · , a(m) = (M − I)m−1 a,
and let
(i)
v0 = [0, a(i) ]t
(2.5)
Then
(2.6)
(1)
vj
∈
/ H and
(m)
vj
(i)
and v1 = [a(i) , 0]t ,
1 ≤ i ≤ m.
are eigenvectors of Tj , j = 0, 1 and
(i)
(i)
(i+1)
Tj vj = vj + vj
(1)
,
1 ≤ i ≤ m − 1, j = 0, 1.
(1)
If we let v0 = v0 and v1 = v1 , then a similar calculation like (2.4) implies that
T0 v1 = T1 v0 again.
Corollary 2.4. Let v0 , v1 be chosen as in the proof of Lemma 2.3, Then
(i)
(i)
(i) T0 v1 = T1 v0
for 1 ≤ i ≤ m.
k−1
(ii) T1 T0 v0 = T0 T1k−1 v1 for k ≥ 1.
(iii) (T0n v0 )1 = (T1n v1 )N = 0
and
(T0n v0 )i = (T1n v1 )i−1 for 2 ≤ i ≤ N .
Proof. (i) and (ii) follows directly from the same calculation as in the proof of the
above lemma. The first identity in (iii) is a consequence of (v0 )1 = (v1 )N = 0 as in
(2.3). For the second identity, if v0 and v1 are 1-eigenvectors of T0 and T1 respectively,
(2.2) implies that
(T0n v0 )i = (v0 )i = (v1 )i−1 = (T1n v1 )i−1 .
For the general case we need only apply
( Pn ¡n¢
(1)
Tjn vj
=
(i+1)
vj
Pm−1 ¡n¢ (i+1)
i=0
i vj
i=0
i
if n < m
if n ≥ m
the regularity of Lp -scaling functions
277
which can be checked directly by using (2.6).
Lemma 2.5. Let v be a 2-eigenvector of (T0 + T1 ) and ṽ = (T0 − I)v. Then
H(ṽ) is a subspace of H. Moreover, if (i) 1 is a simple eigenvalue of T0 and T1 ; or
(ii) H(ṽ) = H, then for v0 , v1 ∈
/ H as defined in Lemma 2.3, there exists a constant
c such that
v = cv0 + h0 = cv1 + h1
for some h0 , h1 ∈ H(ṽ).
Proof. Note that [1, 1, . . . , 1]t is a left 1-eigenvector of T0 , so that (T0 − I)u ∈ H
for every u ∈ Cn . In particular, ṽ = (T0 − I)v must be in H. Also it is easy to show
that H is invariant under T0 and T1 , hence H(ṽ) is a subspace of H. Let v0 and v1
P
P
PN
be as in Lemma 2.3, then a = (v0 )i = (v1 )i 6= 0. Let c = i=1 vi /a, where vi ’s
are the coordinates of v and let
h0 = v − cv0
and h1 = v − cv1 .
By the choice of c, we have h0 , h1 ∈ H which implies case (ii) because H = H(ṽ).
In case (i), we observe that if 1 is a simple eigenvalue of T0 , then T0 − I restricted
on H is bijective; it is hence also bijective on the (T0 − I)-invariant subspace H(ṽ).
Consequently,
ṽ = (T0 − I)v = (T0 − I)(cv0 + h0 ) = (T0 − I)h0
so that h0 must be in H(ṽ). The same proof holds for h1 .
3. Proof of Theorem B. Let f be an Lpc -solution of (1.1) and let v =
[f[0,1) , . . . , f[N −1,N ) ]t be the vector defined by the average of f over the N -subintervals
(see Proposition 2.1), then v is a 2-eigenvector of (T0 + T1 ). Let
X
fn (x) =
(TJ v)χIJ (x)
|J|=n
and let fn be the corresponding real valued function of fn defined on [0, N ].
Lemma 3.1. For n ≥ 1 and ` ≥ 0,
Z
1−2−n
kfn+` (x + 2−n ) − fn+` (x)kp dx
0
=
1
2n+`
X
|J 0 |=`


n
X
X

kTJ (T1 T0i−1 − T0 T1i−1 )TJ 0 vkp  .
i=1 |J|=n−i
Proof. We divide the interval [0, 1 − 2−n ) into 2n − 1 equal subintervals. For each
subinterval, we further divide it into 2` equal parts. In this way we have 2` (2n −1) equal
subintervals with length 2−(n+`) . For each such dyadic interval, we can write down its
binary representation with length 2n+` , say I(j1 ,...,jn ,j10 ,...,j`0 ) . Since it is contained in
[0, 1 − 2−n ), at least one of the j1 , . . . , jn must equal 0. Suppose x ∈ I(j1 ,...,jn ,j10 ,...,j`0 )
with jn−i+1 as the last zero in {j1 , . . . , jn }, i.e., x ∈ I(j1 ,...,jn−i ,0,1,...,1,j10 ,...,j`0 ) , then
278
ka-sing lau and mang-fai ma
x + 2−n ∈ I(j1 ,...,jn−i ,1,0,...,0,j10 ,...,j`0 ) . It follows that
fn+` (x + 2−n ) − fn+` (x)
= Tj1 · · · Tjn−i T1 T0i−1 (Tj10 · · · Tj`0 v) − Tj1 · · · Tjn−i T0 T1i−1 (Tj10 · · · Tj`0 v)
= Tj1 · · · Tjn−i (T1 T0i−1 − T0 T1i−1 )Tj10 · · · Tj`0 v.
Since fn+` (x + 2−n ) − fn+` (x) is a constant function on each dyadic interval of size
2−(n+`) , an integration over the interval [0, 1 − 2−n ) yields the lemma immediately.
We first give a lower bound estimate of k∆2−n f k.
Proposition 3.2. For n ≥ 1,
k∆2−n f kp ≥
2p−1
2n−1
X
kTJ ṽkp .
|J|=n−1
Proof. Fix n ≥ 1 and for any ` ≥ 0,
Z N
k∆2−n fn+` kp =
|fn+` (x + 2−n ) − fn+` (x)|p dx
−2−n
Z
1−2−n
≥
0
=
≥
1
2n+`
2n+`
1
= n
2
|J 0 |=`
X
1
1
≥ n
2
kfn+` (x + 2−n ) − fn+` (x)kp dx


n
X X
X

kTJ (T1 T0i−1 − T0 T1i−1 )TJ 0 vkp 
i=1 |J|=n−i
(by Lemma 3.1)
X
kTJ (T1 − T0 )TJ 0 vkp
|J 0 |=` |J|=n−1
X
|J|=n−1
X
¡1 X
¢
kTJ (T1 − T0 ) `
TJ 0 v kp
2 0
|J |=`
kTJ (T1 − T0 )vkp
(by (2.1))
|J|=n−1
= 2p−1
1
2n−1
X
kTJ ṽkp
(use (T1 − T0 )v = −2ṽ).
|J|=n−1
The assertion now follows by letting ` → ∞.
For the upper bound of k∆h f k, we need an estimation of the integral of |∆h fn (x)|
near the integers k = 0, . . . , N .
Lemma 3.3. Under the same assumptions as in Lemma 2.5, for n > 0 and for
0 < h < 2−n ,
Z
|∆h fn (x)|p dx ≤ 2p h(kT0n h0 kp + kT1n h1 kp )
En
where En =
SN
k=0 [k
− 2−n , k).
the regularity of Lp -scaling functions
279
Proof. Since fn is a constant function on the dyadic intervals of size 2−n , we have
Z
p
|∆h fn (x)| dx =
En
N Z
X
k−2−n
k=0
=
k
N Z
X
k
|fn (x + h) − fn (x)|p dx
|fn (x + h) − fn (x)|p dx
k−h
k=0
N
³
´
X
= h |(T0n v)1 |p +
|(T0n v)i − (T1n v)i−1 |p + | − (T1n v)N |p .
i=2
Recall that v = cv0 + h0 = cv1 + h1 as in Lemma 2.5. Therefore, by Corollary 2.4(iii),
(T0n v)1 = c(T0n v0 )1 + (T0n h0 )1 = (T0n h0 )1
and similarly (T1n v)N = (T1n h1 )N . Also for 2 ≤ i ≤ N , by Corollary 2.4(iii) again,
(T0n v)i − (T1n v)i−1 =c(T0n v0 )i + (T0n h0 )i − c(T1n v1 )i−1 − (T1n h1 )i−1
=(T0n h0 )i − (T1n h1 )i−1 .
We can continue the above estimation:
Z
En
N
³
´
X
|(T0n h0 )i − (T1n h1 )i−1 |p + |(T1n h1 )N |p
|∆h fn (x)|p dx = h |(T0n h0 )1 |p +
p
≤2 h
¡
i=2
[kT0n h0 kp
+ kT1n h1 kp
¢
and complete the proof.
Proposition 3.4. Under the same assumptions as in Lemma 2.5, we have for
0 < h < 2−n ,
k∆h fn kp ≤
´
X
2p+1 ³ X
p
p
kT
h
k
+
kT
h
k
.
J 0
J 1
2n
|J|=n
|J|=n
SN
SN −1
Proof. Let En = k=0 [k − 2−n , k) and Ẽn = [−2−n , N ) \ En = k=0 [k, k + 1 −
2−n ). Since fn is supported by [0, N ], we have
Z
p
k∆h fn k =
N
−2−n
|∆h fn (x)|p dx
Z
Z
|∆h fn (x)|p dx +
=
En
|∆h fn (x)|p dx
Ẽn
:= I1 + I2 .
Lemma 3.3 implies that
I1 ≤ 2p h(kT0n h0 kp + kT1n h1 kp ).
280
ka-sing lau and mang-fai ma
On the other hand, if we write I2 in the vector form, we have
Z 1−2−n
kfn (x + h) − fn (x)kp dx
I2 =
0
=h
n
X
X
kTJ (T1 T0k−1 − T0 T1k−1 )vkp .
k=1 |J|=n−k
From Corollary 2.4(ii) we conclude that
(T1 T0k−1 − T0 T1k−1 )v = T1 T0k−1 (cv0 + h0 ) − T0 T1k−1 (cv1 + h1 )
= T1 T0k−1 h0 − T0 T1k−1 h1 ,
and therefore
I2 ≤ 2p h
n
³X
X
kTJ T1 T0k−1 h0 kp +
k=1 |J|=n−k
≤ 2p h
³ X
X
kTJ h0 kp +
|J|=n
n
X
X
kTJ T0 T1k−1 h1 kp
´
k=1 |J|=n−k
´
kTJ h1 kp .
|J|=n
The lemma then follows from the two estimates of I1 and I2 .
We can now state and prove our main theorem of this section (i.e. Theorem B in
Section 1).
Theorem 3.5. Suppose that either (i) 1 is a simple eigenvalue of T0 and T1 or
PN
(ii) H(ṽ) = { u ∈ CN : i=1 ui = 0 }. If f is a Lpc -solution of (1.1), then
P
ln(2−n |J|=n kTJ ṽkp )
Lipp (f ) = lim inf
.
n→∞
p ln(2−n )
Proof. As a direct consequence of Proposition 3.2, we have
P
ln(2−n |J|=n kTJ ṽkp )
ln k∆2−n f k
Lipp (f ) ≤ lim inf
≤ lim inf
.
n→∞
n→∞
ln(2−n )
p ln(2−n )
To prove the reverse inequality we first observe that k∆h f k ≤ 2kf − fn k + k∆h fn k.
Proposition 2.1 (iii) and Proposition 3.4 imply that
³
´
X
X
X
k∆h f kp ≤ C 2−n
kTJ ṽkp + 2−n
kTJ h0 kp + 2−n
kTJ h1 kp
|J|=n
|J|=n
|J|=n
for some constant C independent of n. Since ṽ, h0 , h1 are all in H(ṽ), we can apply
Lemma 2.2 to have the reverse inequality.
4. Lipp (f ) for some special cases. For the 2-coefficient dilation equation
f (x) = f (2x) + f (2x − 1), the scaling function is χ[0,1) and it is easy to calculate that
Lipp (f ) = 1/p from the definition.
Proposition 4.1. If f is an Lpc -solution of f (x) = c0 f (2x) + c1 f (2x − 1) +
c2 f (2x − 2) with c0 + c2 = 1, c1 = 1, and c0 , c2 6= 0, then
Lipp (f ) =
ln((|c0 |p + |1 − c0 |p )/2)
.
−p ln 2
the regularity of Lp -scaling functions
Proof. In this case,
T0 =
µ
c0
1 − c0
0
1
¶
µ
and T1 =
1
0
c0
1 − c0
281
¶
,
and v = [c0 , c0 − 1]t is a 2-eigenvector of (T0 + T1 ). Then
µ
¶
c0 (c0 − 1)
ṽ = (T0 − I)v =
6= 0.
−c0 (c0 − 1)
Note that ṽ is an c0 -eigenvector of T0 and (1−c0 )-eigenvector of T1 . A straight-forward
calculation yields
n µ ¶
´
1 X
1 ³X n
p
p k
p n−k
kT
ṽk
=
(|c
|
)
(|1
−
c
|
)
kṽkp
J
0
0
2n
2n
k
k=0
|J|=n
³ |c |p + |1 − c |p ´n
0
0
kṽkp .
=
2
This implies that
Lipp (f ) =
ln((|c0 |p + |1 − c0 |p )/2)
.
−p ln 2
We now turn to the 4-coefficient dilation equation
(4.1)
f (x) = c0 f (2x) + c1 f (2x − 1) + c2 f (2x − 2) + c3 f (2x − 3)
with c0 + c2 = c1 + c3 = 1 and c0 , c3 6= 0. We first observe that



1 − c3
c0
c0
0
0
1 − c0
(4.2)
T0 =  1 − c0 1 − c3
c0  , T1 =  c3
0
0
0
c3
1 − c0

0
1 − c3  .
c3
The eigenvalues of (T0 + T1 ) are 2, 1, and (1 − c0 − c3 ), and the 2-eigenvector v is


c0 (1 + c0 − c3 )
(4.3)
v =  (1 + c0 − c3 )(1 − c0 + c3 )  .
c3 (1 − c0 + c3 )
Therefore
(4.4)


c0 (c0 − 1)(1 + c0 − c3 )
ṽ = (T0 − I)v =  −c0 (c0 − 1)(1 + c0 − c3 ) + c3 (c3 − 1)(1 − c0 + c3 )  .
−c3 (c3 − 1)(1 − c0 + c3 )
Note that in Proposition 4.1, the computation can be made easier if ṽ is an eigenvector
of both T0 and T1 . Here we have
Lemma 4.2. Let T0 and T1 be as in (4.2) and let v be the 2-eigenvector of (T0 +T1 )
as in (4.3) and let ṽ = (T0 − I)v. Then ṽ is an eigenvector of both T0 and T1 (not
necessary to the same eigenvalue) if and only if c0 + c3 = 1.
Proof. Suppose c0 + c3 = 1, then c0 = c1 , c2 = c3 , and (4.2) reduces to




c0
0
0
c0
c0
0
T0 =  1 − c0
c0
c0  , T1 =  1 − c0 1 − c0
c0  ,
0
1 − c0 1 − c0
0
0
1 − c0
282
ka-sing lau and mang-fai ma
and ṽ = [−2c20 c3 , 2c20 c3 − 2c0 c23 , 2c0 c23 ]t 6= 0. By a direct calculation, ṽ is a c0 eigenvector of T0 and (1 − c0 )-eigenvector of T1 .
Conversely, suppose ṽ is an eigenvector of both T0 and T1 . Let u0 = [0, 1, −1]t
and u1 = [1, −1, 0]t , then ṽ = au0 + bu1 where a and b isPdetermined by (4.4). By
using u0 and u1 as a basis of the subspace H = {u ∈ C3 :
ui = 0}, we can rewrite
T0 , T1 (restricted on H) and ṽ as follows:
µ
¶
µ
¶
µ ¶
1 − c0 − c3 c3
c3
0
a
(4.5)
T0 =
, T1 =
, and ṽ =
.
0
c0
c0 1 − c0 − c3
b
Note that T0 has c0 and 1 − c0 − c3 as eigenvalues while T1 has c3 and 1 − c0 − c3
as eigenvalues. We claim that ṽ is an c0 -eigenvector of T0 . For otherwise, ṽ is an
(1−c0 −c3 )-eigenvector of T0 , then b must be zero and ṽ = [a, 0]t . But this contradicts
to the assumption that ṽ is an eigenvector of T1 . Similarly, ṽ must be a c3 -eigenvector
of T1 . Hence,
(T0 + T1 )ṽ = (c0 + c3 )ṽ.
There are only three choices of the eigenvalues of T0 + T1 : 2, 1 or 1 − c0 − c3 . By a
direct check we conclude that c0 + c3 = 1 is the only allowable case.
In view of Lemma 4.2 we can use the same technique as in Proposition 4.1 to
prove the next proposition
Proposition 4.3. If f is an Lpc -solution of (4.1) with the additional assumption
that c0 + c3 = 1, then
(4.6)
Lipp (f ) =
ln((|c0 |p + |1 − c0 |p ))/2
.
−p ln 2
In Figure 1, we draw the graphs of some scaling functions satisfying the assumption in the above proposition and their Lp -Lipschitz exponents. Note that if
Lipp (f ) = 1 for all 1 ≤ p < ∞, then f is differentiable almost everywhere and the
derivative is in Lp for all 1 ≤ p < ∞. This is the case for c0 = 0.5 and is obvious from
the graph of the corresponding scaling function. For the graph of c0 = 1.125, we see
that Lipp (f ) is undefined for p > 6. Indeed f ∈
/ Lp (R), for p > 6, making use of the
criterion in Theorem A.
We conclude this section by giving a formula of Lipp (f ) with the coefficients
satisfying c0 + c3 = 12 instead of c0 + c3 = 1. It includes Daubechies scaling function
√
√
D4 which corresponds to c0 = (1 + 3)/4, c3 = (1 − 3)/4. This formula has been
obtained in [DL3] using a different method and assuming in addition that 21 < c0 < 34 .
Here, we need an estimation on the product of two non-commutative matrices.
Lemma 4.4. Let β0 , β1 ∈ R. Let
µ
¶
µ
¶
1 0
1
0
P0 =
and P1 =
.
0 β0
β0 β1
For any multi-index J = (j1 , j2 , . . . , jn ), we let PJ = Pj1 · · · Pjn . Then
µ
¶
1
0
PJ =
λJ µJ
where λJ = β0 (j1 + j2 βj1 + · · · + jn (βjn−1 · · · βj1 )) and µJ = βjn βjn−1 · · · βj1 . Let
the regularity of Lp -scaling functions
1.0
283
1.0
c0 = 0.625
c0 = 0.5
0.8
0.6
0.6
f(x)
f(x)
0.8
0.4
0.4
0.2
0.2
0.0
0.0
0.5
1.0
1.5
2.0
2.5
0.0
0.0
3.0
0.5
1.0
1.5
2.0
x
x
Figure 1a
Figure 1b
2.5
3.0
3
1.0
c0 = 1.125
c0 = 0.875
2
0.8
1
f(x)
f(x)
0.6
0
0.4
-1
0.2
-2
0.5
1.0
1.5
2.0
2.5
0.0
3.0
0.5
1.0
1.5
2.0
x
x
Figure 1c
Figure 1d
c0 = 0.5
1.0
c0 = 0.625
0.8
Lipp(f)
0.0
0.0
0.6
c0 = 0.875
0.4
0.2
c0 = 1.125
0.0
1
2
3
4
5
6
p
Figure 1e
7
8
9
10
2.5
3.0
284
ka-sing lau and mang-fai ma
γ = (|β0 |p + |β1 |p )/2. Then
X
2−n
|µJ |p = γ n
2−n
and
|J|=n
X
|λJ |p ≤ C np max{1, γ n }
|J|=n
for some constant C > 0 independent of n.
Proof. The explicit form of the product PJ can easily be shown by induction. For
the second part of the lemma, the first identity follows from
X
X
2−n
|βjn · · · βj1 |p = γ n .
|µJ |p = 2−n
j1 ,...,jn =0,1
|J|=n
For the second identity we observe that
µ
µ X
¶ p1
X
|λJ |p
= |β0 |
¯
¯p ¶ p1
n
X
¯
¯
¯j1 +
¯
j
(β
·
·
·
β
)
i ji−1
j1 ¯
¯
j1 ,...,jn =0,1
|J|=n
i=2
µ
n µ
X
(n−1)/p
≤ |β0 | 2
+
i=2
X
p
¶ p1 ¶
|ji (βji−1 · · · βj1 )|
j1 ,...,jn =0,1
(by Minkowski inequality)
µ
¶
n
X
¡
¢
(n−1)/p
p
p (i−1)/p
= |β0 | 2
+
|β0 | + |β1 |
i=2
≤ |β0 | 2(n−1)/p
≤ |β0 | n 2
n
X
(γ 1/p )i−1
i=1
(n−1)/p
max{1, (γ 1/p )n }.
The last identity now follows.
Proposition 4.5. If f is the Lpc -solution of (4.1) with the additional assumption
that c0 + c3 = 21 , then
½
¾
ln((|c0 |p + | 12 − c0 |p )/2)
(4.7)
Lipp (f ) = min 1,
.
−p ln 2
Proof. In this case, (4.2) reduces to


1
c0
0
0
2 + c0
1



c0
T0 = 1 − c0 2 + c0
and T1 = 12 − c0
1
1 − c0
0
0
2 − c0
c0
1 − c0
0
1
2
1
2

0
+ c0  .
− c0
Note that h = [1, −2, 1]t is a c0 -eigenvector of T0 and also a ( 12 − c0 )-eigenvector of
T1 . It is clear that
µ
¶n
X
1
2−n
kTJ hkp = 2−n |c0 |p + | − c0 |p khkp .
2
|J|=n
Since h ∈ H(ṽ), by Lemma 2.2, we have
Lipp (f ) ≤
ln((|c0 |p + | 12 − c0 |p )/2)
.
−p ln 2
the regularity of Lp -scaling functions
285
Next observe that u = [0, 1, −1]t is a 12 -eigenvector of T0 and T1 u = 12 u + c0 h.
Therefore, by using u and h as a basis of the subspace H(ṽ), we can rewrite T0 , T1 ,
restricted on H (= H(ṽ) in this case), and ṽ, as
µ1
¶
µ 1
¶
µ ¶
0
0
a
2
2
T0 =
, T1 =
, and ṽ =
,
0 c0
b
c0 12 − c0
where a = − 14 and b = 23 c0 (c0 − 1)( 21 + 2c0 ). Let β0 = 2c0 and β1 = 1 − β0 . For
J = (j1 , j2 , . . . , jn ), we have TJ = 21n PJ , so that
kTJ ṽkp = |a2−n |p + |2−n (aλJ + bµJ )|p ,
where λJ and µJ are defined as in Lemma 4.4. This implies kTJ ṽkp ≥ |a2−n |p and
Lipp (f ) ≤ 1 by Theorem 3.5. Hence
¾
½
ln((|c0 |p + | 12 − c0 |p )/2)
(4.8)
Lipp (f ) ≤ min 1,
.
−p ln 2
On the other hand,
kTJ ṽkp ≤ |a2−n |p + 2p |a2−n λJ |p + 2p |b2−n µJ |p .
By Lemma 4.4, we have
X
2−n
kTJ ṽkp ≤ C np 2−pn max{1, ((|β0 |p + |β1 |p )/2)n }
|J|=n
½
¾
¡
¢n
= C np max 2−pn , (|β0 /2|p + |β1 /2|p )/2
.
Consequently we have the reverse inequality of (4.8) and completes the proof.
In figures 2a–e we again sketch some Lpc -scaling functions from Proposition 4.5
(c0 + c3 = 12 ) and their Lp -Lipschitz exponents Lipp (f ) of p. The case for c0 = 0.25
corresponding to χ[0,1] ∗ χ[0,1] , it is differentiable and hence Lipp (f ) = 1 for all p. The
case corresponding to c0 = 0.683... is the Daubechies scaling function D4 . From the
picture of Lipp (f ), D4 has Lp -derivative for 1 ≤ p < 2. It is known that for p = 2, D4
is differentiable almost everywhere but the derivative
is not in L2 . Also it is known
√
that the Hölder exponent of D4 is 2 − ln(1 + 3)/ ln 2, which is the same number as
the formula in the proposition when p → ∞.
5. Lipp (f ) when p is a positive even integer. The computation of Lipp (f )
in Section 4 depends on the existence of an eigenvector of both T0 and T1 (which
may be associated with different eigenvalues). This technique cannot be used for the
general case. In this section we show that if p is a positive even integer, then Lipp (f )
is related to the spectral radius of a matrix Wp whose entries are induced from the
coefficients of the dilation equation. For simplicity, we only give the construction of
Wp for the 4-coefficient dilation equation. It is not hard to extend this to the case
with more coefficients.
In view of Theorem 3.5, we will first develop a simple expression for the sum
P
2−n |J|=n kTJ ṽkp for p a positive even integer. Let [0, 1, −1]t and [1, −1, 0]t be a
P
basis of H = {u ∈ C3 :
ui = 0}. Then T0 and T1 can be written as in (4.5). Let
e0 = [1, 0], e1 = [0, 1]. For a fixed u = [α, β]t ∈ H(ṽ) (to be determined later), we
286
ka-sing lau and mang-fai ma
0.5
1.0
c0 = 0.25
c0 = 0.683... (D4)
0.8
0.4
0.6
0.3
f(x)
f(x)
0.4
0.2
0.2
0.0
0.1
-0.2
-0.4
0.0
0.0
0.5
1.0
1.5
2.0
2.5
3.0
0.0
1.0
1.5
2.0
x
Figure 2a
Figure 2b
2.5
3.0
3
1.0
c0 = 1.125
c0 = 0.875
2
0.8
1
0.6
0
f(x)
0.4
0.2
-1
-2
0.0
-3
-0.2
-4
-0.4
0.0
0.5
1.0
1.5
2.0
2.5
-5
0.0
3.0
0.5
1.0
1.5
2.0
2.5
x
x
Figure 2c
Figure 2d
c0 = 0.25
1.0
c0 = 0.683... (D4)
0.8
Lipp(f)
f(x)
0.5
x
0.6
c0 = 0.875
0.4
0.2
c0 = 1.125
0.0
1
2
3
4
5
6
p
Figure 2e
7
8
9
10
3.0
the regularity of Lp -scaling functions
define the vector an with the i-th entry by
X
(an )i =
(e0 TJ u)p−i (e1 TJ u)i ,
287
i = 0, . . . , p.
|J|=n
If p is an even integer, then
X
X
kTJ ukp =
(|e0 TJ u|p + |e1 TJ u|p )
|J|=n
|J|=n
(5.1)
= (an )0 + (an )p = |(an )0 | + |(an )p |.
Note that a0 = [αp , αp−1 β, . . . , αβ p−1 , β p ]t . If we let d = 1 − c0 − c3 , we have, in
view of (4.5),
and hence
(an+1 )i =
e0 T0 = de0 + c3 e1 ,
e1 T0 = c0 e1 ,
e0 T1 = c3 e0 ,
e1 T1 = c0 e0 + de1 ,
X
(e0 TJ u)p−i (e1 TJ u)i
|J|=n+1
=
X
|J|=n
=
X
(e0 T0 TJ u)p−i (e1 T0 TJ u)i +
X¡
|J|=n
+
(e0 T1 TJ u)p−i (e1 T1 TJ u)i
|J|=n
¢p−i
(de0 + c3 e1 )TJ u
(c0 e1 TJ u)i
X¡
¢p−i
(c3 e0 )TJ u
(c0 e0 + de1 TJ u)i
|J|=n
=
à p−i µ
X p − i¶
X
|J|=n
+
`
`=0
X ³
¶
p−i µ
X
p−i
`=0
`
dp−i−` (e0 TJ u)p−i−` c`3 (e1 TJ u)`
p−i
cp−i
3 (e0 TJ u)
à i µ ¶
´ X
i
`=0
|J|=n
=
!
ci0 c`3 dp−i−` (an )i+` +
`
Summarizing the above, we have
Proposition 5.1. For any integer p ≥ 1,
defined by
 ¡p−i¢ i j−i p−j

j−i c0 c3 d


i
(Wp )ij = ci0 dp−i + cp−i
3 d


 ¡ i ¢ i−j p−i j
j c0 c3 d
ci0 (e1 TJ u)i
¢
!
i−` `
ci−`
d (e1 TJ u)`
0 (e0 TJ u)
i µ ¶
X
i
`=0
¡
`
p−i `
ci−`
0 c3 d (an )` .
let Wp be the (p + 1) × (p + 1) matrix
for 0 ≤ i < j ≤ p
for i = j
for 0 ≤ j < i ≤ p
where d = 1 − c0 − c3 . Then
an+1 = Wp an = Wpn+1 a0
288
ka-sing lau and mang-fai ma
where a0 = [αp , αp−1 β, . . . , αβ p−1 , β p ]t . In particular if p is an even integer, then
X
kTJ ukp = [1, 0, 0, · · · , 0, 1]Wpn a0 .
|J|=n
(L)
(U )
(L)
(U )
The matrix Wp can be written as Wp = Wp + Wp , where Wp and Wp are
the lower and upper triangular part of Wp , in a very symmetric manner. For example,
¡2¢ 2 
 ¡2¢ 2 ¡2¢

 ¡0¢ 2
c3
0
0
1 c3 d
0 d
0 c3
¡1¢
¡1¢2
¡1¢



 ¡ 1¢
(U )
(L)
;
W2 =  0 c0 c3
0  , W2 =  0
0 c0 d
1 c0 c3 
1 c3 d
¡0¢ 2
¡2¢ 2
¡2¢ 2 ¡2 ¢
0
0
0 c0
1 c0 d
2 d
0 c0
 ¡0¢ 4

0
0
0
0
0 c3
 ¡1¢c c3 ¡1¢c3 d
0
0
0 

 0 0 3
1 3
 ¡ 2¢

¡
¢
¡
¢
(L)
2
2 2 2
2
2
2

W4 =  0 c0 c3
0
0 
,
1 c0 c3 d
2 c3 d
¡3¢ 2
¡3¢
¡3¢ 3

 ¡ 3¢ 3
2
 0 c0 c3
0 
1 c0 c3 d
2 c0 c3 d
3 c3 d
¡4¢ 4
¡4¢ 3
¡4¢ 2 2 ¡4 ¢ 3 ¡4 ¢ 4
0 c0
1 c0 d
2 c0 d
3 c0 d
4 d
¡4¢ 3
¡4¢ 4 
 ¡4 ¢ 4 ¡4 ¢ 3 ¡4¢ 2 2
c3
0 d
1 c3 d
2 c3 d
3 c3 d
¡3¢ 3 ¡3¢
¡3¢ 2
¡3¢4 3 
 0
2
c0 c3 d
c0 c3 d 3 c0 c3 

0 c0 d

¡1 2¢ 2 2 ¡22¢ 2
¡2¢ 2 2 
(U )

W4 =  0
0
c0 c3 
.
0 c0 d
1 c0 c3 d
¡1¢ 3
¡12¢ 3 

 0
0
0
c0 c3 
0 c0 d
¡1 0¢ 4
0
0
0
0
0 c0
Recall from basic linear algebra that if ρ(A) is the spectral radius of an N × N
matrix A, then ρ(A) = max{|λ| : λ is an eigenvalue of A} and
(5.2)
lim kAn k1/n = ρ(A).
n→∞
Let λ be any eigenvalue of A, and Eλ = {u ∈ CN : (A − λI)m u = 0 for some m > 0},
then CN = Eλ ⊕ Z for some A-invariant subspace Z of CN . We say that u has
a component in Eλ if u = uλ + z with uλ 6= 0. It is clear that if u ∈ CN has a
component in Eλ , then there is a constant C > 0 such that kAn uk ≥ C|λ|n for all
n > 0.
Lemma 5.2. Let λ be the eigenvalue of Wp such that |λ| = ρ(Wp ) and let Eλ
be defined as above. Suppose dim H(ṽ) = 2. Then there exists u = αb0 + βb1 ∈
H(ṽ), where b0 = [0, 1, −1]t , b1 = [1, −1, 0]t , such that H(u) = H(ṽ) and the
corresponding a0 = [αp , αp−1 β, . . . , αβ p−1 , β p ]t has a component of Eλ . For such u,
there is a constant C > 0 such that
X
C ρ(Wp )n ≤
kTJ ukp
for all n > 0.
|J|=n
Proof. We choose p + 1 vectors ui = αi b0 + βi b1 , i = 0, . . . , p, such that H(ui ) =
H(ṽ) and
αi βj − αj βi 6= 0,
for i 6= j.
the regularity of Lp -scaling functions
289
Then the corresponding vectors
γi = [αip , αip−1 βi , . . . , αi βip−1 , βip ]t ,
0≤i≤p
form a basis of Cp+1 because the matrix with the vectors γi ’s as rows is a Vandermonde
matrix

 p
α0 α0p−1 β0 · · · α0 β0p−1 β0p
 αp αp−1 β1 · · · α1 β p−1 β p 
1 
1
1
 1
A= .
..
..
.. 
 ..
.
.
. 
αpp
αpp−1 βp
···
αp βpp−1
βpp
Q
and detA = 0≤j,k≤p (αj βk − αk βj ) 6= 0. Hence, one of the γi ’s has a component of
Eλ . Let u be the corresponding ui and the first part of the lemma follows. To prove
the second part, we observe that for any J with |J| = n and 0 ≤ j ≤ p,
|e0 TJ u|p−j |e1 TJ u|j ≤ max{|e0 TJ u|p , |e1 TJ u|p }
≤ |e0 TJ u|p + |e1 TJ u|p .
Hence
|(an )j | =
X
|e0 TJ u|p−j |e1 TJ u|j
|J|=n
≤
X
|e0 TJ u|p +
|J|=n
=
X
X
|e1 TJ u|p = |(an )0 | + |(an )p |
|J|=n
kTJ uk
p
(by (5.1))
|J|=n
Pp
P
It follows that kan k1 = j=0 |(an )j | ≤ p |J|=n kTJ ukp . Since a0 has a component
of Eλ , there exists a constant C > 0 such that
C ρ(Wp )n ≤ kWpn a0 k1 = kan k1 ≤ p
X
kTJ ukp .
|J|=n
For the 4-coefficient dilation equation in (4.1), it is easy to check that dim H(ṽ) =
0 if and only if (c0 , c3 ) ∈ {(0, 0), (1, 0), (0, 1), (1, 1)}, and the solutions are characteristic functions ([LW, Lemma 3.3]). Hence Lipp (f ) = 1/p. Also dim H(ṽ) = 1 if and
only if c0 + c3 = 1 (Lemma 4.2), and in Proposition 4.3 we have given a formula of
Lipp (f ) for this case. It remains to consider the case H(ṽ) = 2, which will complete
all the cases for all 4-coefficient scaling functions.
Theorem 5.3. Consider the 4-coefficient dilation equation in (4.1) with the assumption that dim H(ṽ) = 2. For p a positive even integer, the equation has a (unique)
Lpc -solution f if and only if ρ(Wp )/2 < 1, and in this case
Lipp (f ) =
ln(ρ(Wp )/2)
.
−p ln 2
290
ka-sing lau and mang-fai ma
Proof. Note that for any u ∈ H(ṽ) and for any ² > 0, we have for large n
X
kTJ ukp = |(an )0 | + |(an )p |
(by (5.1))
|J|=n
≤ kan k1 = kWpn a0 k1
≤
(by Proposition 5.1)
kWpn k ka0 k1
≤ ka0 k1 (ρ(Wp ) + ²)n .
(by (5.2))
If we choose u ∈ H(ṽ) as in Lemma 5.2, combining with Theorem A in Section 1, we
have
Pthe first conclusion. The second assertion follows from Lemma 2.2, the estimation
of |J|=n kTJ ukp from above and Lemma 5.2.
Figure 3 shows the domain of (c0 , c3 ) for the existence of Lpc -solutions for even
integers using the above criterion ρ(Wp )/2 < 1. The curves are ρ(Wp )/2 = 1 corresponds to p = 2, 4, 6, 10, 20, and 40. Note that when p → ∞ the limit is the
triangular region which is the approximate region plotted in [H] for the existence of
continuous 4-coefficient scaling functions using the joint spectral radius. However, we
are not able to prove this assertion yet, i.e., limp→∞ Lipp (f ) is the Hölder exponent.
Also we do not have a criterion for the existence of an L∞
c -solution.
Figure 4 is the graph of Lip4 (f ) plotted against the (c0 , c3 )-plane. It shows the
overall picture of Lip4 (f ) for the 4-coefficient case. It looks similar to the graph of
Lip2 (f ) plotted in [LMW].
We remark that if c0 > 0, c3 > 0, and 1 − c0 − c3 > 0, then T0 and T1 in (4.5) are
non-negative matrices. Also the vector u in Lemma 5.2 can be chosen to be a positive
vector. Hence (5.1) still holds if p is a positive odd integer. Consequently, we have
Proposition 5.4. Consider the 4-coefficient dilation equation (4.1) with c0 > 0,
c3 > 0, and 1 − c0 − c3 > 0. Suppose dim H(ṽ) = 2, then for p a positive integer,
(5.3)
Lipp (f ) =
ln(ρ(Wp )/2)
.
−p ln 2
Without such positivity assumption on the coefficients, the expression in (5.3)
does not necessarily give Lipp (f ) for p odd integers. For example Figure 5a is the
graph of −ln(ρ(Wp )/2)/(p ln 2), p = 1, · · · , 10, of the scaling function corresponding
to c0 = 0.5 and c3 = −0.4. The points bounce up and down but Lipp (f ) should
be convex in that region. Figure 5b corresponds to Daubechies scaling function D4
(c3 < 0). On this graph, the points are obtained by −ln(ρ(Wp )/2)/(p ln 2) while the
curve is Lipp (f ) given by (4.7). It shows that for even integer p, they coincide. For
odd integer p, the values obtained by (5.3) are different to Lipp (f ) but surprisely
close (see Table 1). Also when p → ∞, in our numerical and graphical experiments,
the values obtained from −ln(ρ(Wp )/2)/(p ln 2), p odd integers, seems to converge to
Lipp (f ) rather rapidly.
Finally we remark that for the dilation equation with N +1 (N > 3) coefficients, if
dim H(ṽ) = 1 then ṽ is an eigenvector of both T0 and T1 , say T0 ṽ = aṽ and T1 ṽ = bṽ.
Then the same technique as in the proof of Proposition 4.1 yields
Lipp (f ) =
ln((|a|p + |b|p )/2)
−p ln 2
for 1 ≤ p < ∞. For the case dim H(ṽ) ≥ 2, we can use a similar method to that in
the regularity of Lp -scaling functions
291
1.5
1.0
c3
0.5
0.0
-0.5
-1.0
-1.0
-0.5
0.0
0.5
1.0
1.5
c0
Figure 3
1.0
Lip 4(f)
0.5
0.0
1.5
1.0
0.5
-0.5
-0.5
0.0
0.0
c0
0.5
c3
-0.5
1.0
1.5
-1.0
Figure 4
Proposition 5.1 to obtain a (p + 1)N −2 × (p + 1)N −2 square matrix Wp , and Theorem
5.3 and Proposition 5.4 will still hold for dim H(ṽ) ≥ 2.
292
ka-sing lau and mang-fai ma
c0 = 0.5
c3 = -0.4
1.0
f = D4
1.0
0.8
Lipp (f)
0.9
0.6
0.4
0.8
0.7
0.2
0.6
1
2
3
4
5
6
7
8
9
10
1
2
3
4
5
6
7
p
p
Figure 5a
Figure 5b
p
3
5
7
9
Lipp (f )
0.874185416
0.749617426
0.692852392
0.661125656
8
9
10
value obtained
by (5.3)
0.892690635
0.750414497
0.692893269
0.661127939
Table 1
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